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Algebra and Applications Test 2

Question : 1 of 40

Marks: +1, -0

(1 + w^{r-s})^{-1} + (1 + w^{s-r})^{-1}

Solution: 👈: Video Solution

Concept:

Use substitution to simplify the symmetric expression.

Explanation:

Let

x = w^{r-s}

Then

w^{s-r} = x^{-1}

The expression becomes

\frac{1}{1+x} + \frac{1}{1+x^{-1}}

Simplify the second term:

\frac{1}{1+x^{-1}} = \frac{x}{1+x}

So the total is

\frac{1}{1+x} + \frac{x}{1+x} = \frac{1+x}{1+x} = 1

Since

w

is a cube root of unity,

x \neq -1

, so the denominators are not zero.

Answer:

The value is

1

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